The "vector" part of this problem is to break down where the camel is walking into two components: a component which is only concerned with north-south travel, and a component which is only concerned with east-west travel.
You have the initial angle that the camel is walking: 15° west of due south; and it travels 23km. Using trigonometry, you can determine how much W of that 23km is just going westward, and how much S is just going southward. (Note that these two components don't add to 23km! Using Pythagoras' theorem, you should be able to see that W2 + S2 = (23km)2 .) So first, determine W and S. This will give you the position (call it C) of the camel after it's first blundering attempt to go to B.
Note that the usual convention is to describe west as being (negative east), and south as being (negative north). That is, because the camel is heading south and west, you should be able to describe its position in terms of a negative distance north, and separately a negative distance east.
The camel then starts going due north; it should be easy to find out where it ends up after that leg of the trip, because all of that leg is invested in north-south travel; you can just add that to the north-south component of the camel's position, to get its position D after the second leg.
Now you know where the camel is after the second leg of it's haphazard odyssey. Construct a right-angle triangle with the corners at B and D, and with the two shorter sides parallel to the x and y axes (that is, horizontal and vertical; or north-south and east-west). You should be able to apply Pythagoras' Theorem to find out how far away D is from B; that is, how far the camel has to travel to finally arrive where it was supposed to go. And for the direction, use trigonometry to find out the angle that it must travel, relative to east (which is the positive direction of the x axis).
Good luck!
It is a common mistake for people to confuse
The arctan of $x$
with
The solution to $\tan \theta = x$
For every value of $x$, the equation has infintiely many solutions, and it has two different solutions with $\theta \in [0^\circ, 360^\circ)$ (or in any interval of length $360^\circ$ you choose). Similar statements are true for each of the trig functions.
We can see the dangers of making this mistake in your work: although you compute a vector $\vec{D}$ in the third quadrant, the polar form you computed is for a vector in the first quadrant!
The $\tan$ function satisfies $\tan \theta = \tan(\theta + n 180^\circ)$ (where $n$ can be any integer). Within the interval $[0^\circ, 360^\circ)$, the other solution that you missed is thus $41^\circ + 180^\circ$, which does give an angle in the third quadrant.
Some methods of calculation provide an alternative to $\arctan$ to avoid this issue; rather than computing an angle via $\arctan(y/x)$, they offer a different version of $\arctan$ that takes two arguments, and always gives an angle in the correct quadrant for the vector $(x,y)$. e.g.
Best Answer
If we are starting from point $(0,0)$ then after the first step we are in $(0,53)$ and after the second we got to $(0 + 45\sin \alpha, 53 + 45 \cos \alpha)$. $\sin \alpha = \frac{\sqrt 3}{2}, \cos \alpha = \frac 12$, so the resulting point is $(\frac{45\sqrt3}{2},\frac{151}{2})$ and, as our starting point was $(0,0)$, this is also the displacement vector.